Find a domain in which $F(z)$ is holomorphic and satisfies $F'(z)=\frac{1}{z^2+1}$ Given a function $$f(z)=\frac{1}{z^2+1}$$the goal is to find a holomorphic primitive of the function $F(z)$.
I have approached this problem by laurent expanding $f$ in $\Omega=\{|z|<1\} $ and then integrating. My approach is as follows:
$$f(z)=\sum_{n\ge0}(-1)^n\ z^{2n}$$
$$F(z)=a_0+\sum_{n\ge0}\frac{(-1)^n}{2n+1}\ z^{2n+1}$$
Is this a valid approach i.e. is $F$ holomorphic in $\Omega$? Is there something more elegant?
 A: HINT: Note that $$\frac{1}{z^2 +1 } = \frac{i}{2} \left(\frac{1}{z+i} - \frac{1}{z-i}\right). $$
Consider the complex logarithms $\log (z+i)$  and $\log(z-i)$ with appropriated domains. 
A: It is a valid approach to see if you want $F$ to be holomorphic only on $\Omega$, the open unit disk.
By your method, you found that the function 
$$F(z)=\sum_{n=0}^\infty \frac{(-1)^nz^{2n+1}}{2n+1}$$
is a valid primitive to your function $f$. Since every power series is analytic on their disk of convergence, $F$ is holomorphic on $\Omega$. This power series does equal $\arctan z$ inside the unit disk.
However, you cannot find a holomorphic function on a domain that contain a simple closed curve around $i$ or around $-i$ individually like $\mathbb{C}$ or $\mathbb{C}\setminus{\{-i,i\}}$ such that its derivative is $f$. 
To show this, let $\gamma$ be a simple closed curve around $i$ of winding number $1$, then by the residue theorem, it is easily shown that
$$\int_{\gamma}f(z)\, \mathrm{d}z=\pi$$
However, this integral should be zero if $\gamma$ was included in the domain of a primitive of $f$.
